Question

Solve the given set of equations by applying L-U decomposition method. The upper triangular matrix U should include 1's on its diagonal. 1.5x1+2.4x2-2.1x3=-1.86 2.4x1+7.04x2-0.16x3=-2.656 1.4x1+3.44x2-3.16x3=-4.496

          Solve the given set of equations by applying L-U decomposition method. The upper triangular matrix U should include 1's on its diagonal.
1.5x1+2.4x2-2.1x3=-1.86
2.4x1+7.04x2-0.16x3=-2.656
1.4x1+3.44x2-3.16x3=-4.496
        
Solve the given set of equations by applying L-U decomposition method. The upper triangular matrix U should include 1's on its diagonal.
1.5x1+2.4x2-2.1x3=-1.86
2.4x1+7.04x2-0.16x3=-2.656
1.4x1+3.44x2-3.16x3=-4.496

Added by Edward C.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Solve the given set of equations by applying LU decomposition method. The upper triangular matrix U should include 1's on its diagonal. 1.5x + 2.4x^2 - 2.1x^3 = -1.86 2.4x + 7.04x^2 - 0.16x^3 = -2.656 1.4 + 3.44x^2 - 3.16x^3 = -4.496
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00:01 In this question we have been given a linear system we have to factor the coefficient matrices into lu decomposition with lii is equals to 1 this is for all i so we will write the given system in the form of ax equals to b where a is the coefficient matrix so i will write the matrix as 1 2 4 1 i am not writing the equation here you can refer to the questions this is 2 minus 2 minus 1 3 2 then we have 3 1 minus 1 minus 2 1 3 minus 1 minus 1 since we have to factorize it i will let a is equals to lu where we know that l is a lower triangular matrix so i will get on the diagonal everywhere it is 1 and above it we have zeros everywhere and this will be l 2 1 l 3 1 l 3 2 this is l 4 1 l 4 2 l 4 3 lii is equals to 1 that means each of the diagonal entries is 1 and then we write the upper triangular matrix u11 u12 u13 u14 next is 0 u22 u23 u24 then 0 0 u33 u34 and here everywhere 0 this will be u44 so this is how we write a is equals to lu now if i multiply and compare multiply and compare both sides what we are going to get is following things u11 equals to 1 u12 comes out to be equal to 2 u13 is equals to 4 and u14 we get as minus 1 so this is if we compare row r1 next if i compare row r2 that is a second row on both the side i will get l 2 1 u11 equals to 2 u11 will know from that we get l 2 1 equals to 2 similarly i will get l 2 1 u12 plus u22 equals to minus 1 which implies l u22 will be minus 5 next is l 2 1 times u13 plus u23 i get it to be 3 which implies u23 is equals to minus 5 also i get that u14 times l 2 1 plus u24 equals to 2 which implies…
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